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प्रश्न
The following table shows the marks scored by 80 students in an examination:
| Marks | 0 – 5 | 5 – 10 | 10 – 15 | 15 – 20 | 20 – 25 | 25 – 30 | 30 – 35 | 35 – 40 |
| No. of students |
3 | 10 | 25 | 49 | 65 | 73 | 78 | 80 |
उत्तर
Let us choose a = 17.5, h = 5, then `d_i = x_i – 17.5 and u_i = (x_i− 17.5)/5`
Using step-deviation method, the given data is shown as follows:
| Marks | Number of students (cf) |
Frequency `(f_i)` |
Class mark `(x_i)` |
`d_i = x_i` – 17.5 |
`u_i = (x_i−17.5)/ 5` |
`(f_i u_i)` |
| 0 – 5 | 3 | 3 | 2.5 | -15 | -3 | -9 |
| 5 – 10 | 10 | 7 | 7.5 | -10 | -2 | -14 |
| 10 – 15 | 25 | 15 | 12.5 | -5 | -1 | -15 |
| 15 – 20 | 49 | 24 | 17.5 | 0 | 0 | 0 |
| 20 – 25 | 65 | 16 | 22.5 | 5 | 1 | 16 |
| 25 – 30 | 73 | 8 | 27.5 | 10 | 2 | 16 |
| 30 – 35 | 78 | 5 | 32.5 | 15 | 3 | 15 |
| 35 – 40 | 80 | 2 | 37.5 | 20 | 4 | 8 |
| Total | `Ʃ f_i = 80` | `Ʃ f_i u_i = 17` |
The mean of the given data is given by,
x = a+ `((Ʃ_i f_i u_i )/(Ʃ_i f_i)) xx h`
= `17.5 + (17/80) xx 5`
= 17.5 + 1.06
= 18.56
Thus, the mean marks correct to 2 decimal places is 18.56.
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